Consider the marks,out of $100$,obtained by $51$ students of a class in a test,given in the table below. Draw a frequency polygon corresponding to this frequency distribution table.

Marks	Number of students
$0-10$	$5$
$10-20$	$10$
$20-30$	$4$
$30-40$	$6$
$40-50$	$7$
$50-60$	$3$
$60-70$	$2$
$70-80$	$2$
$80-90$	$3$
$90-100$	$9$
Total	$51$

(N/A) To draw a frequency polygon,we first represent the data using a histogram. We mark the mid-points of the tops of the rectangles as $B, C, D, E, F, G, H, I, J, K$ respectively.
Since the first class is $0-10$,to close the polygon,we extend the horizontal axis in the negative direction to include an imaginary class-interval $(-10) - 0$ with frequency $0$. We mark the mid-point of this interval as the starting point. The line segment from this point joins to the first mid-point $B$. Similarly,we take an imaginary class-interval $100-110$ with frequency $0$ and mark its mid-point $L$. The last mid-point $K$ is joined to $L$.
The resulting figure $OABCDEFGHIJKL$ is the required frequency polygon.
Alternatively,frequency polygons can be drawn using class-marks. The class-mark is calculated as:
$\text{Class-mark} = \frac{\text{Upper limit} + \text{Lower limit}}{2}$